I have a question about voltage drop. I know how to use D3 Table, I can figure out the voltage drop for a two wire circuit No-problem. I don't know how to figure out the voltage drop for a three phase circuit though. In the old code books it was easy because on Table D5 they used to give you the -CM Area- in the 2002 code book they don't. The old formula: CM=(K)(I)(L)(1.73)
---------------
VD

Table D5 has taken away the CM Area.
So how do you find the voltage drop in a three phase circuit using the 2002 CEC?
For example three-phase (no neutral) going to a motor controller.
Sorry for all the questions but I thought that I would take advantage of all the expertise.

Hey is anybody out there? Hey Tony could you help me with my question?

Thanks

All that I am looking for is if I was to use the 2002 CEC only would I be able to find the voltage drop in the three phase circuit. I don't think I can. I think that I would require some other book most likely a manufacturers pamplet with info on the specific wire. I'm not sure though maybe I'm not looking in the right place in the code book that's why I asking the question. I am positive that people have calculated voltage drops in three phase circuits before. I am just curious on how they do it with just the code book.
Thanks

Robbie,

Here is the formula in calculating Voltage Drops in Three Phase Circuits:


Three Phase Formulas as Follows:

%VD,3PH= KFactorxAmpsxLength of CCT in Metres
Phase to Phase Supply Voltage-volts


Single Phase Formulas as Follows:

%VD,1PH= KFactorx1.15xAmpsxLength in Metres
Phase to Neutral Supply Voltage

If you are using the OESC 2002'look under Table D3.

For other nominal voltages, multiply the distances in metres by the other nominal voltage (in volts) and divide by 120.


Example on use of Table:

Consider a two conductor circuit of No. 12 AWG copper NMD90 carrying 16 A at nominal 240 V under maximum ambient of 30°C. The maximum run distance from the centre of distribution to the load without exceeding a 3% voltage drop is:

Maximum run length for No. 12 AWG, 16 A, 1% voltage drop at nominal 120 V from Table is: 6.1 m

Distance Correction Factor to be used is:

From Table 2, allowable ampacity for 2
conductor No. 12 AWG NMD90 (90°C rating per Table 19) is 20 A. The given current is 16 A or 80% of the allowable ampacity.

The Distance Correction Factor to be used, from Note (3), 90°C row, 80% column, is 1.00.


The maximum run length is:

6.1 m x 3% x 1.00 x 240v = 37m
120v

Beyond this distance a larger size of conductor is required, ie, No. 10 AWG (30 A allowable ampacity) beyond 37 m up to and including 62 m.

9.7 ,m x 3% x 1.06 x 240v = 62m
120

Tony Moscioni
Electrical Inspector
Electrical Safety Authority

Thanks Tony, Is it possible to calculate the voltage drop of a three phase circuit using only the 2002 CEC?